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I want to calculate the work done by friction if the length $L$ of uniform rope on the table slides off. There is friction between the cord and the table with coefficient of kinetic friction $\mu_k$.

$$ W = \int F \cdot d \vec{s}$$ I think it would be: $$ W_{fr} = \frac M L g \int_{0}^{L} dx$$

But the solutions (which could be mistaken) say: $$ dW_{fr} = \mu_k \frac M L g \, x \, dx$$ which is then integrated.

Should there be an $x$ in the integral? I don't think there should be because you are summing up over an infinitesimal displacement $dx$ and the force of friction is not proportional to the displacement at any instant (I think).

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  • $\begingroup$ Do you know that your expression is dimensionally incorrect? $\endgroup$ – ABC Apr 6 '13 at 3:27
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Let $x$ denote the length of the rope that is on the table, then $$ m(x) = \frac{M}{L}x $$ is the mass of the rope on the table. It follows that the force of friction on the rope on the table is $$ f(x) = \mu_k m(x) g = \mu_k\frac{M}{L}xg $$ if the rope moves an amount $dx$ then the work done by friction is $$ dW = f(x)dx = \mu_k\frac{M}{L}gx dx $$

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